Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.8 - The Divergence Theorem and a Unified Theory - Exercises 16.8 - Page 1025: 1

Answer

$0$

Work Step by Step

We know that $ div F=\dfrac{\partial P}{\partial x}i+\dfrac{\partial Q}{\partial y}j $ So, $ div \space F=\dfrac{\partial}{\partial x}[\dfrac{-y}{(x^2+y^2)^{1/2}}]+\dfrac{\partial}{\partial y} [\dfrac{x}{(x^2+y^2)^{1/2}}]$ and $ div \space F=\dfrac{2xy}{2}(x^2+y^2)^{-3/2}-\dfrac{2yx}{2}(x^2+y^2)^{-3/2}=0$
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